 #1
agoogler
 81
 0
Homework Statement
I'm trying to solve the following problem :
In △ABC, coordinates of B are (−3,3). Equation of the perpendicular bisector of side AB is 2x+y−7=0. Equation of the perpendicular bisector of side BC is 3x−y−3=0. Mid point of side AC is E(11/2,7/2). Find AC^{2}.
Homework Equations

The Attempt at a Solution
By solving 3x−y−3=0 and 2x+y−7=0 I find that the intersection of perpendicular bisectors is at (2,3) .
Then using the two points (2,3) and (11/2,7/2), I get the equation of perpendicular bisector of AC as y=x/7+19/7.
So the slope of AC is 7 and then using point slope form , y−7/2=−7(x−11/2) Thus the equation of line AC is y=42−7x .
Similarly equation of line BC is y=2−x/3 .
So AC and BC intersect at (6,0).
By using the fact that E is the midpoint of AC, I find Coordinates of A as (5,7).
So the distance between A and C is 5√2, and AC^{2}=50.
But this answer is wrong and the correct answer is 74 ( I checked the answer sheet) .
What have I done wrong ?